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I'm interested in Buchberger's criterion for determining if G={g_1,....,g_n} is a Grobner basis for the ideal it generates. In the procedure, I consider the S-polynomial S(g_i,g_j) and check if it has remainder zero upon division by G in any order.

My question is if Buchberger's criterion holds if one use a different order of G for each i,j rather than a fixed but arbitrary choice of order of G for all i,j.

In other words, could G be falsely Grobner even if S(g_i,g_j) has remainder 0 by dividing by G in different orders depending on (i,j)?

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  • $\begingroup$ Could you be a bit more clear? The property of being a Groebner basis depends on an ordering. So what is it that you are asking? Is it "Suppose that for each pair $(i,j)$ there is an order for which $S(g_i,g_j)$ has remainder 0. Is it true that for some order $G$ is a Groebner basis?" - or something else? $\endgroup$
    – Vladimir Dotsenko
    Commented Jul 24, 2013 at 2:30
  • $\begingroup$ When I mean ordering, I don't mean the term order, but rather the order of the generators by which you execute the multivariate division algorithm. BTW, I've looked at Cox-Little-O'Shea and I think the answer is that one can use different orders (in the latter sense), but just want to confirm. $\endgroup$
    – Simpleperson
    Commented Jul 24, 2013 at 3:43
  • $\begingroup$ Oh I see. That was a very confusing way to formulate it. So you mean: "Is it enough to check that $S(g_i,g_j)$ has remainder 0 for some multivariate long division procedure?", right? $\endgroup$
    – Vladimir Dotsenko
    Commented Jul 24, 2013 at 15:21

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Indeed, it is enough to check that each $S(g_i,g_j)$ can be reduced to zero in some way. Indeed, if you trace carefully the proof of Buchberger's criterion, you will see that one only needs the existence of a representation of $S(g_i,g_j)=\sum_{k}a_{ijk}g_k$ with each leading term of $a_{ijk}g_k$ being less than or equal to the leading term of $S(g_i,g_j)$, which is precisely what you can get from the existence of long division that produces remainder $0$.

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